ELEC207 Coursework (v6: 1 February 2022)
Page 1
ELEC207 Coursework: Design of a Stable Martian Segway (“Experiment 81”)
Prof Simon Maskell 1 February 2022
Module ELEC207
Coursework name Experiment 81
Component Weight 25% = 3.75 Credits
Semester 2
HE Level 5
Lab location Third floor EEE building
Work Individually
Timetabled time Check Canvas announcements for lab module
Suggested Private Study time 16h
Assessment method Individual report
Submission Format On line submission - Canvas
Late Submission Standard University Penalties
Resit Opportunity By arrangement
Marking Policy Numerical mark
Anonymous marking Yes
Feedback Canvas
Subject of relevance Control Engineering
AHEP Learning Outcomes LO1
This coursework component1 of ELEC207 relates to Part B, “Control”, and focuses on the
content up to and including lecture 10 of Part B of the module. The mark you will receive for
the coursework constitutes 25% of your mark for ELEC207 and is intended to enable you to
demonstrate your understanding of how to:
• use the position of poles to demonstrate whether systems are stable;
• use your knowledge of control to define a controller that ensures that a system is
stable;
• use root locus to ensure the closed-loop time-response has specific properties;
• use Simulink to validate that the closed-loop time-response is as expected;
• explain your work in a clear and concise fashion.
You are expected to make use of the lecture notes and explicit references to numbered
lectures are therefore included in this document. The demonstrators associated with the lab
are available to support you in undertaking this coursework. Queries can also be submitted
via the discussion board for ELEC207 on Canvas.
The mark you will receive (out of a total of 40 marks) will quantify the following aspects of
your write-up:
• Demonstration of your understanding of ELEC207 (75% and out of 30 marks);
• Clarity of exposition (25% and out of 10 marks).
The marking descriptors are provided in the appendix.

1 Assessment of ELEC207 has previously included Experiment 81, which may result in some legacy references
to experiment 81 in documentation that has not yet been updated to reflect the change. This coursework
takes the place of experiment 81.
ELEC207 Coursework (v6: 1 February 2022)
Page 2
Any emboldened text in a box herein implies that a specific response should be included in
your write-up with the number in brackets indicating the number of marks associated with
that component of the write-up. Failure to include such a response is liable to result in you
obtaining fewer marks than would have been the case otherwise.
The assessment of the assignment is intended to be sufficiently straightforward that a
diligent student should be able to achieve a pass mark of 40% but sufficiently challenging
that achieving a first (ie 70% or above) requires deep understanding of the subject matter.
To aid you in understanding how challenging each mark is to obtain, marks are annotated
with E for Easy, M for Moderate and H for Hard: 8 of the marks are deemed to be easy; 14
are deemed moderate; 8 are deemed hard.
You should submit your coursework on or before the deadline announced by the lab
coordinator (check Canvas announcements).
Plagiarism and collusion or fabrication of data is always treated seriously and action
appropriate to the circumstances is always taken. The procedure followed by the University
in all cases where plagiarism, collusion or fabrication is suspected is detailed in the
University’s Policy for Dealing with Plagiarism, Collusion and Fabrication of Data, Code of
Practice on Assessment, Category C, available on:
https://www.liverpool.ac.uk/media/livacuk/tqsd/code-of-practi...
ELEC207 Coursework (v6: 1 February 2022)
Page 3

  1. Mathematical Modelling
    A Segway (as shown in figure 1) is a physical system that can be modelled as an inverted
    pendulum.
    Figure 1: A Segway
    More specifically, we will assume that the pendulum can be approximated as a point mass,
    of mass m, at a distance, l, and at a (small) angle, θ(t), defined clockwise from vertical.
    Gravity is assumed to act downwards and exert an acceleration of g. A motor provides a
    torque, T(t). The actuator that converts the Torque control signal to the physical Torque can
    be assumed to have a gain of unity. The angular acceleration of the pendulum can then be
    approximated as being defined by:
    𝑔
    𝑙
    θ(t) + T(t) = 𝑚
    𝑑
    2θ(t)
    𝑑𝑡
    2
    Our task as an engineer designing the controller for the Martian Segway is to optimise the
    time-response to changes to an input, X(t), which defines the desired values for θ(t). We will
    consider this design criterion to be achieved if the settling time is equal to ts.
    To ensure your Martian Segway is as unique as your coursework, please assume that:
    • l is the day of the month when you were born (where l is in metres);
    • m is the month of the year when you were born (where m is in kg);
    • ts is the year when you were born divided by 250 (where ts is in seconds).
    Please define the values for l, m and ts that you will use for your coursework. [1E]
    Now derive the transfer function, 𝑯(𝒔) = 𝜽(𝒔)⁄𝑻(𝒔), of the Segway in terms of l, m and
    g. [1E]
    Using your values for l and m along with g=3.711 ms-2
    , write the transfer function with
    the denominator and numerator of your transfer function in polynomial form. [1E]
    Calculate the position of the poles for your Segway and plot the poles on the complex
    plane. [1E]
    ELEC207 Coursework (v6: 1 February 2022)
    Page 4
  2. Validating that the Open-loop System is Unstable
    With no controller, the system is believed to be unstable. We can use Simulink to validate
    this. Open MATLAB R2020a or later (as available using Apps Anywhere or the Remote
    Teaching Centre Service2
    ) and then click Simulink. Information is available online (at:
    https://uk.mathworks.com/help/simulink) on how to use Simulink (please also check the
    supporting materials available in your Canvas lab module page). Here we provide a brief
    description of what is needed for our purposes:
  3. Press “Blank Model”;
  4. Click the “Library Browser” button, ;
  5. Search for a step function by entering “Step” into the box where it says “enter search
    term”;
  6. Drag and drop the “Step” block onto the Simulink Editor;
  7. Click the “Library Browser” button again;
  8. Search for “Transfer Fcn” and drag and drop the “Transfer Fcn” block onto the
    Simulink Editor. Be careful to select the continuous-time transfer function block
    (parameterised by s) and not the discrete-time transfer function block (parameterised
    by z);
  9. Double click the “Transfer Fcn” block on the Simulink Editor and enter the
    polynomials that you calculated above. Be careful that any coefficients of zero need
    to be explicitly entered (ie coefficients of [1 0 2] describe s
    2+2 but coefficients of [1 2]
    describe s+2);
  10. Click on the triangle on the right of the “Step” block such that an arrow appears;
  11. Connect the output signal from the “Step” block to the input signal for the “Transfer
    Fcn” block;
  12. Add an “Outport” block;
  13. Connect the output signal from the “Transfer Fcn” block to the input signal for the
    “Outport” block;
  14. Selecting the signal that connects the “Transfer Fcn” block to the input for the
    “Outport” block such that it turns blue;
  15. Click “Add viewer” (under “Simulation” and “Prepare”) and select “Scope”;
  16. Press “Run” (under “Simulation” and “Simulate”). You should now see a graph
    showing the time-response of your Segway to a unit-step.
  17. Press “File” -> “Print” -> “Microsoft Print to PDF” -> “OK” and save the time-response
    to a convenient location.
    Insert a picture of the time-response of your Segway to the unit-step. [2E]
    Comment on whether this time-response indicates that the open-loop system is
    stable. [1M]
  18. Ensuring that the Closed-loop System is Stable Using PID Control
    We will now add a cascade controller, as explained in lecture 7. We will use a combination of
    Proportional, Integral and Derivative control, ie a PID controller. The transfer function of a
    PID controller is:

2 You can also download a copy of Matlab/Simulink from the University’s Computer Services Department
website at https://www.liverpool.ac.uk/csd/software/software-downloads/#...
ELEC207 Coursework (v6: 1 February 2022)
Page 5
𝐶(𝑠) = 𝐾𝑝 +
𝐾𝐼
𝑠

  • 𝐾𝐷𝑠 =
    𝐾𝑝𝑠 + 𝐾𝐼 + 𝐾𝐷𝑠
    2
    𝑠
    where Kp, KI and KD are respectively the proportional, integral and derivative control
    constants. We know that the closed-loop transfer function for a negative feedback system
    with open-loop transfer function of C(s)H(s) and that relates the input, X(s), to the output,
    θ(s), is:
    𝜃(𝑠)
    𝑋(𝑠)
    =
    𝐶(𝑠)𝐻(𝑠)
    • 𝐶(𝑠)𝐻(𝑠)
      Write the closed-loop transfer function for your Segway in terms of Kp, KI and KD as a
      ratio of polynomials in s. Ensure that the highest order term in s in the denominator
      has a coefficient of unity. [3M]
      To ensure that the system is stable, we start by placing the poles at s=-1, s=-2 and s=-3.
      What is the characteristic polynomial that would result in these pole positions? [1M]
      By equating the coefficients in the closed-loop transfer function’s denominator and
      this characteristic function, deduce values for Kp, KI and KDwhich will ensure that the
      closed-loop system is stable. [3M]
  • Validating That the Closed-loop System is Stable
    Simulink includes a “Subtract” block and a “PID” block. Use these blocks (with the
    parameters of the PID controller that you have defined) to simulate the time-response from a
    stable closed-loop system.
    Insert a picture of the time-response of your closed-loop system to the unit-step. [2M]
  • Optimising the Time-Response Using Root Locus
    The open-loop system has poles and zeros that are the union of those associated with the
    plant and those associated with the PID controller.
    Calculate the positions of the open-loop zeros (ie the zeros of 𝑪(𝒔)𝑯(𝒔)) for the values
    of l, m, Kp, KI and KD that you have used. [1M]
    Recall that the PID controller has a pole at the origin and the Segway has poles that you
    have calculated above.
    State the positions of the open-loop poles (ie the poles of 𝑪(𝒔)𝑯(𝒔)) for the values of l
    and m that you have used. [2E]
    We are going to use the root locus to choose the open-loop gain to achieve the timeresponse we want. More specifically, we are going to move the closed-loop poles along the
    root locus such that we can achieve the settling time that you have defined at the start of this
    coursework assignment.
    As explained in lecture 6, the settling time for a generalised second-order system with
    natural frequency, β, and damping coefficient, ω, is 𝑡𝑠 ≈ 4⁄β𝜔. As explained in lecture 5, the
    real part of the location of the poles of a generalised second-order system is 𝑅𝑒(𝑠) = −β𝜔.
    We therefore want to ensure that the dominant poles (the poles that are closest to the
    ELEC207 Coursework (v6: 1 February 2022)
    Page 6
    imaginary axis) are located such that 𝑅𝑒(𝑠) = − 4 𝑡 ⁄ 𝑠
    . We can understand how to achieve
    this by sketching the root locus for the open-loop system with the compensator.
    Sketch the root locus for 𝑪(𝒔)𝑯(𝒔) and identify the points on the root locus that are
    such that 𝑹𝒆(𝒔) = − 𝟒⁄𝒕𝒔
    . [3M]
    We now want to calculate the value of the open-loop gain that will ensure that the closedloop poles are such that we achieve the desired time-response.
    Write the open-loop transfer function, 𝑪(𝒔)𝑯(𝒔), as a ratio of polynomials in 𝒔. [1H]
    If the numerator of the open-loop transfer function is 𝑍(𝑠) and the denominator of the openloop transfer function is 𝑃(𝑠), ie 𝐶(𝑠)𝐻(𝑠) = 𝑍(𝑠)⁄𝑃(𝑠), then the closed-loop poles occur
    when 𝑃(𝑠) + 𝐾𝑍(𝑠) = 0.
    Write 𝑷(𝒔) + 𝑲𝒁(𝒔) = 𝟎 as a polynomial in 𝒔 involving 𝑲. [1H]
    To identify the value of 𝐾, we need to re-express this polynomial as a polynomial in 𝑠̃= 𝑠 +
  • 𝑡 ⁄ 𝑠
    . We can achieve this by substituting 𝑠 = 𝑠̃− 4 𝑡 ⁄ 𝑠
    (using your value for 𝑡𝑠
    ) and
    simplifying.
    Write 𝑷(𝒔̃) + 𝑲𝒁(𝒔̃) = 𝟎 as a polynomial in 𝒔̃ involving 𝑲. [1H]
    As explained in lecture 9, we can then use Routh-Hurwitz to deduce the value of 𝐾 that is
    such that 𝑠̃is on the imaginary axis (ie when 𝑅𝑒(𝑠) = − 4 𝑡 ⁄ 𝑠 and so we achieve our desired
    time-response). We achieve this by choosing 𝐾 to be such that there is a row of zeros.
    Complete a Routh table for 𝑷(𝒔̃) + 𝑲𝒁(𝒔̃). Deduce the value of 𝑲 that is such that
    𝑹𝒆(𝒔) = − 𝟒⁄𝒕𝒔
    [3H]
  • Validating the Response of Optimised System
    You can now use Simulink to simulate the time-response from the closed-loop system with
    the gain you have chosen, 𝐾. Note that there is a “Gain” block that you may find useful.
    Insert a picture of the time-response of your improved closed-loop system to the unitstep. [2H]
  • Further Directions for Private Study (Not Assessed)
    Should you find this coursework assignment interesting and wish to continue to work on
    designing a Martian Segway, it might be interesting to consider:
    1) Using a compensator (eg a PD controller) to ensure that the overshoot adheres to
    some design criterion;
    2) Using a further compensator (eg a PI controller) to ensure that the steady-state error
    is reduced.
    For the avoidance of doubt, your mark for the coursework will not be affected by whether you
    design these compensators.
    ELEC207 Coursework (v6: 1 February 2022)
    Page 7
    Marking descriptors
    Demonstration of your
    understanding of ELEC207 (75%)
    Exposition and structure of the
    report (25%)
    90-100%
    ‘Outstanding’
    Total coverage of the experiment aims,
    objectives and task set. An exceptional
    demonstration of knowledge and understanding,
    appropriately grounded in theory and relevant
    literature. Outstanding research and academic
    content.
    Extremely clear exposition. Excellently logical
    structure. Excellent presentation, only the
    most insignificant errors. Scientific
    dissemination.
    80-89%
    ‘Excellent’
    As ‘Outstanding’ but with some minor
    weaknesses in knowledge. Original and novel
    aspects presented but not fully developed.
    As ‘Outstanding’ but with some minor
    weaknesses in structure, logic and/or
    presentation.
    70-79%
    ‘Very Good’
    Full coverage of the task set. A very good
    demonstration of knowledge and understanding
    but with some modest gaps. A very good
    grounding in theory.
    Very clear exposition. Satisfactory structure.
    Very good presentation, largely free of
    grammatical and other errors. All compulsory
    sections present.
    60-69%
    ‘Comprehensive’
    As ‘Very Good’ but with some gaps in knowledge
    and understanding and/or gaps in theoretical
    grounding.
    As ‘Very Good’ but with some weaknesses in
    exposition and/or structure, a few more
    grammatical and other errors.
    50-59%
    ‘Competent’
    Covers most of the task set. Patchy knowledge
    and understanding with a limited grounding in
    literature.
    Competent exposition and structure.
    Competent presentation but some significant
    grammatical and other errors. Minor errors
    in labelling graphs and figures.
    40-49%
    ‘Adequate’
    As ‘Competent’ but patchy coverage of the task
    set and more weaknesses and/or omissions in
    knowledge and understanding. Just meets the
    threshold level.
    As ‘Competent’ but with more weaknesses in
    exposition, structure, presentation and/or
    errors. Just meets the threshold level.
    35-39%
    ‘Compensatable
    fail’
    Some parts of the set task are likely to have been
    omitted. Major gaps in knowledge and
    understanding. Some significant confusion. Very
    limited grounding. Falls just short of the
    threshold level.
    Somewhat confused and limited exposition.
    Confused structure. Some weaknesses in
    presentation and some serious grammatical
    and other errors. Falls just short of the
    threshold level.
    20-34%
    ‘Deficient’
    As ‘Compensatable Fail’ but with more serious
    weaknesses in presentation and/or grammar.
    Falls substantially below the threshold level.
    As ‘Compensatable Fail’ but with more
    serious weaknesses in presentation and/or
    grammar. Falls substantially below the
    threshold level.
    0-20%
    ‘Extremely weak’
    Largely confusing exposition and structure. Many
    serious grammatical and other errors.
    Largely confusing exposition and structure.
    Many serious grammatical and other errors.
    WX:codinghelp

温文尔雅的生姜
1 声望0 粉丝